$\overline{BC} = 16$ $\overline{AC} = {?}$ $A$ $C$ $B$ $?$ $16$ $ \sin( \angle BAC ) = \dfrac{4}{5}, \cos( \angle BAC ) = \dfrac{3}{5}, \tan( \angle BAC ) = \dfrac{4}{3}$
$\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the opposite side and need to solve for the adjacent side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{16}{\overline{AC}} $ Since we have already been given $\tan( \angle BAC )$ , we can set up a proportion to find $\overline{AC}$ $ \tan( \angle BAC ) = \dfrac{4}{3} = \frac{16}{\overline{AC}}$ Simplify. $\overline{AC} = 12$